Is entropy the volume in phase space of energy E or LESS than E?

[nonequilibrium]

Hello,

I thought the statistical definition of entropy for an isolated system of energy E (i.e. microcanonical ensemble) was S=k \ln \Omega where \Omega is the volume in phase space of all the microstates with energy E.

However, if you take a look here http://en.wikipedia.org/wiki/Equipartition_theorem#The_microcanonical_ensemble
there is the line

\textrm{... Similarly, $\Sigma(E)$ is defined to be the total volume of phase space where the energy is less than $E$ ...} \textrm{By the usual definitions of statistical mechanics, the entropy $S$ equals $k_B \log \Sigma(E)$ ...}

so they use the volume in phase space where energy < E instead of the surface where energy = E. Do these notions coincide? I would think they'd conflict. Why do they say "by the usual definitions", I'm confused.

 

[nonequilibrium]

I'm finding a source (Huang, Statistical Mechanics, 2nd edition, p134) that states S = k \log \Omega and S = k \log \Sigma are indeed equivalent up to a constant dependent of N. The reason for that, I don't seem to get, as the text is a bit too advanced for me atm.

In a way I'm willing to accept the equivalency (as it would clear up my problem), but there's one thing that bothers me: take for example a state of a certain system such that if you lower the energy, entropy goes up (think of a system with bounded energy), doesn't the \Sigma(E) (= the volume in phase space where energy < E) definition make this behavior impossible, because surely (by definition) E_1 &lt; E_2 \Rightarrow \Sigma(E_1) &lt; \Sigma(E_2) \Rightarrow S(E_1) &lt; S(E_2)?

What am I overlooking?